llm tutors
Great results today using Claude as a tutor. Setup:
- A Coursera class Differential Equations for Engineers
- Emacs
- Claude
- gptel
- LaTeX
I am solving exercises in an org-mode notebook with wonderful formula preview, and Claude is able to double-check or proof-read those quite easily.
A sample session follows.
\[ \lim_{w \to w0} \frac{ \cos{wt} - \cos{w_0 t} }{ w_0^2 - w^2 } = ? \]
We have:
\[ w_0^2 - w^2 = (w_0 - w) (w_0 + w) \]
\[ \lim_{w \to w0} \frac{ \cos{wt} - \cos{w_0 t} }{ w_0^2 - w^2 } = \frac{ \lim_{w \to w0} \frac{\cos{wt} - \cos{w_0 t}}{w_o - w} }{2w} \]
\[ \lim_{w \to w0} \frac{\cos{wt} - \cos{w_0 t}}{w_o - w} = ? \]
How would you solve this? Give me a hint.
Looking at your limit:
\[\lim_{w \to w_0} \frac{\cos{wt} - \cos{w_0 t}}{w_0 - w}\]
Hint: This is the definition of a derivative!
Specifically, if you let \(f(w) = \cos(wt)\), then this limit is asking for \(-f'(w_0)\) (note the negative sign because of \(w_0 - w\) in the denominator instead of \(w - w_0\)).
So you need to:
- Find the derivative of \(\cos(wt)\) with respect to \(w\)
- Evaluate it at \(w = w_0\)
- Don't forget the negative sign
What's \(\frac{d}{dw}[\cos(wt)]\)?
Using chain rule:
\[ f(g(x))' = f'(g(x)) g'(x) \]
\[ \frac{d}{dw}[\cos(wt)] = - t sin(wt) \]
\[ -f'(w_0) = t sin(w_0 t) \]
So we get:
\[ \lim_{w \to w_0} \frac{\cos{wt} - \cos{w_0 t}}{w_0 - w} = t sin(w_0 t) \]
And finally:
\[ \lim_{w \to w0} \frac{ \cos{wt} - \cos{w_0 t} }{ w_0^2 - w^2 } = \frac{t sin(w_0 t)}{2w_0} \]
Is this correct?